Question 1133555
<br>
If AD and BC are tangents to a circle at the endpoints of diameter AB, and if AC and BD intersect at a point on the circle, then AD and BC are perpendicular to diameter AB and are on the same side of AB.<br>
Let E be the point of intersection of AC and BD; E is on the circle.  Let F be on AB with EF perpendicular to AB.<br>
Using similar triangles (or other methods) it can be determined that EF = (AD*BC)/(AD+BC) = 144/25.<br>
Also using similar triangles, it can be determined that AF:FB = 9:16.<br>
So let AF=9x anf FB=16x; then once again similar triangles tell us that<br>
{{{AF/EF=EF/FB}}}
{{{9x/(144/25)=(144/25)/16x}}}
{{{144x^2 = (144/25)^2}}}
{{{x^2 = 144/625}}}
{{{x = 12/25}}}<br>
Then the length we are looking for, AB, is AF+FB = 25x = 12.